Question

Решить логарифмическое неравенство:

Answer 1

$x*3^{log_{_{frac{1}{9}}}(16x^4-8x^2+1)} textless frac{1}{3} \ODZ:16x^4-8x^2+1 textgreater 0 \Tochki peresecheniya: 16x^4-8x^2+1=0 \C TEOPEMbI BUETA: x^2=frac{4}{16} \x^2=frac{1}{4} \x=pmfrac{1}{2} \Esli x=-1, to 16x^4-8x^2+1=16-8+1=9 \Esli x=0, to 16x^4-8x^2+1=1 \Esli x=1, to 16x^4-8x^2+1=9 (funktsiya parnaya) \ODZ: xneq pmfrac{1}{2}$
$x*3^{log_{_{frac{1}{9}}}(16x^4-8x^2+1)} textless frac{1}{3} \x*3^{{log_{_{3^{-2}}}}(16(x-frac{1}{4})^2)}}}} textless frac{1}{3} \x*3^{-frac{1}{2}log_{_3}(4(x-frac{1}{4}))^2} textless frac{1}{3} \x*3^{{-log_{_3}}(4(x-frac{1}{4}))}} textless frac{1}{3} \frac{x}{3^{{log_{_3}}(4x-1)}}} textless frac{1}{3} \frac{x}{4x-1} textless frac{1}{3} \frac{x}{4x-1}-frac{1}{3} textless 0 \frac{3x-4x+1}{3(4x-1)} textless 0 \frac{1-x}{12x-3} textless 0 \imeem sovokupnost' sistem: \ left { {{1-x textless 0} atop {12x-3 textgreater 0}} right. ili left { {{1-x textgreater 0} atop {12x-3 textless 0}}$
$\ left { {{1-x textless 0} atop {12x-3 textgreater 0}} right. ili left { {{1-x textgreater 0} atop {12x-3 textless 0}} right. \ left { {{x textgreater 1} atop {12x textgreater 3}} right. left { {{x textless 1} atop {12x textless 3}} right. \left { {{x textgreater 1} atop {x textgreater frac{3}{12}}} right. left { {{x textless 1} atop {x textless frac{3}{12}}} right. \left { {{x textgreater 1} atop {x textgreater frac{1}{4}}} right. left { {{x textless 1} atop {x textless frac{1}{4}}} right. \x textgreater 1 ili x textless frac{1}{4}$
$Uchitivaem ODZ: xneq pm frac{1}{2} \OTBET: xin {(-infty;-frac{1}{2});(-frac{1}{2};frac{1}{4});(1;+infty)$
Графическое решение на изображении

Answer 2

А как это записать всё?